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Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free Lie algebra. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

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  • $\begingroup$ No combinatorists out there? I think this is a basic enough question! $\endgroup$
    – Chitrabhanu
    Commented Feb 5, 2013 at 17:14
  • $\begingroup$ Yes it is basic but up to my knowledge very little is known on these structure constants. Do you know more now ? $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 7, 2015 at 9:18
  • $\begingroup$ If anybody's still following this question: Mr Duchamp, you say "little is known on these structure constants". Could you tell us what is known (and where to find it)? More specifically, I'm searching for a formula/algorithm that gives me $[B(a),B(w)] = \sum_u \alpha_u B(u)$ with $a$ a letter, $w$ and the $u$'s Lyndon words, and the $\alpha_u$ numbers (which I understand will be integers). $\endgroup$
    – David Vercauteren
    Commented Nov 17, 2016 at 8:14
  • $\begingroup$ @DavidVercauteren I am now far from closely monitoring this subject, sorry. What you can remark is that, to each Lyndon word is associated a binary tree (see Lothaire, Combinatorics on Words Chapter 5.3). $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 21, 2018 at 3:24

2 Answers 2

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Yes it is of this form because any Lie polynomial "begins" by a Lyndon word, in particular, the least monomial of $B(l)$ is $l$. Then $$ B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v\qquad \mathbf{(LB)} $$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$.

Computationally

  1. $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
  2. $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
  3. But, $mn<nm$ because $mn$ is Lyndon and then $$ [B(m),B(n)]=mn+[sb1]-nm-[sb2]=mn+[sb3] $$ where the square bracket $[sb3]$ is a linear combination of monomials that are greater than $mn$ or $nm$

Hence all (monomials of $[sb3]$) are greater than $mn$ which is the form you required.

One can say a little bit more From property $\mathbf{(LB)}$, and the fact that $[B(m),B(n)]$ is a multihomogeneous Lie polynomial, one has ($\underline{w}$ being the commutative image of $w$) $$ [B(m),B(n)]=B(mn) + \sum_{mn<l\atop l\ \mathrm{Lyndon};\ \underline{l}=\underline{mn}} \gamma_{m,n}^{(l)}B(l) $$
the coefficients $\gamma_{m,n}^{(l)}$ are the structure constants of the free Lie algebra w.r.t. the Lyndon-Sirsov basis (they are integers, universal, i.e. characteristic free) and, up to my knowledge, their combinatorics is widely unknown.

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As far as I can see, your statement is equivalent to Lemma 3.5 in http://arxiv.org/pdf/0804.1254.pdf .

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  • $\begingroup$ Yes, equivalent up to a reversal (just to clarify in order that readers be not misled). In the statement, a Lyndon word is the (strict) minimum of a (primitive) conjugacy class as in Reutenauer's book Free Lie algebras (which I believe has become standard convention now) whereas in the paper you cite (Lemma 3.5), a Lyndon-Sirsov word is the (strict) maximum of a (primitive) conjugacy class as was used in the former "école de Lille" around Gérard Jacob. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 8, 2015 at 7:02
  • $\begingroup$ See, for instance <a href="sciencedirect.com/science/article/pii/S0012365X99001247" target="_blank">sciencedirect.com/s<wbr>cience/article/pii/…>. It seems that the question of structure constants is still widely open. $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 8, 2015 at 7:11
  • $\begingroup$ This, of course, does not withdraw any parcel of the merit of Sirsov who was the first to point out these words (rediscovered as explained in Bokut's paper independently by Fox and Lyndon). $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 8, 2015 at 7:22
  • $\begingroup$ @DuchampGérardH.E. Yes, I don't think much understood about structure constants. I feel however that the OP was asking a question about a sort of triangularity property with some structure constants, which is indeed very easy. $\endgroup$
    – Vladimir Dotsenko
    Commented Jun 8, 2015 at 7:41
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    $\begingroup$ OK, if it is this, any of us can answer $\endgroup$
    – Duchamp Gérard H. E.
    Commented Jun 8, 2015 at 8:06

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